TOPIC 10
Ray model of light
10.1 Overview
10.1.1 Module 3: Waves and Thermodynamics
Ray model of light
Inquiry question: What properties can be demonstrated when using the ray model of light?
Students:
•• conduct a practical investigation to analyse the formation of images in mirrors and lenses via reflection
and refraction using the ray model of light (ACSPH075)
•• conduct investigations to examine qualitatively and quantitatively the refraction and total internal reflection of light (ACSPH075, ACSPH076)
•• predict quantitatively, using Snell’s Law, the refraction and total internal reflection of light in a variety of
situations
•• conduct a practical investigation to demonstrate and explain the phenomenon of the dispersion of light
•• conduct an investigation to demonstrate the relationship between inverse square law, the intensity of light
and the transfer of energy (ACSPH077)
•• solve problems or make quantitative predictions in a variety of situations by applying the following relationships to:
c
–– nx = – for the refractive index of medium x, vx is the speed of light in the medium
vx
–– n1sin (i) = n2sin (r) (Snell’s Law)
1
–– sin (ic) = – for the critical angle ic of medium x
nx
–– I1r21 = I2r22 – to compare the intensity of light at two points, r1 and r2
FIGURE 10.1 Rainbows are the result of refraction, reflection and dispersion of light
rays by fine water droplets in the air.
TOPIC 10 Ray model of light 215
10.2 What is light?
10.2.1 Electromagnetic waves
10.2.2 Sources of light
Light sources of different kinds can be
classified as being either luminous or illuminated. Luminous bodies are those that
emit electromagnetic radiation in the visible part of the spectrum either as a result
of chemical processes or because they are
incandescent. Incandescent objects glow
because they are very hot and, the hotter
they are, the more light they produce. The
Sun’s incandescence is the result of the
enormous heat generated by the thermonuclear reactions within its interior. On a
much smaller scale, the tiny particles of
hot carbon produced in a candle flame
provide incandescent light, and the tungsten filament in a light bulb can produce
light as a result of becoming white hot
when electric charges move through it.
216 Jacaranda Physics 11
FIGURE 10.2 The electromagnetic spectrum.
1023
10–14
1022
10
21
Gamma rays
10
20
1019
10
18
Frequency (Hz)
1015
1014
10–12
10–11
X-rays
Ultraviolet
VISIBLE
10–8
10–6 1μm
10–5
Infra-red
1012
Microwaves
and
mobile phones
10–3
Short radio waves
10–1
Television and
FM radio
1
101
AM radio
102
1010
109
108
107
106
10
104
103
102
10
10–4
10–2 1 cm
103
5
Long radio waves
500
10–7
1013
1011
450
10–10
10–9 1 nm
1017
1016
400 nm
10–13
Wavelength (m)
Light is the term commonly used to
describe electromagnetic radiation; more
specifically, we tend to use it to describe
only a small part of the electromagnetic
spectrum known as the visible spectrum.
Electromagnetic radiation travels in the
form of transverse waves. However, unlike
mechanical waves such as sound, earthquake tremors and pond ripples, electromagnetic waves do not need a medium to
travel in; in fact, they slow down when
travelling in any physical medium apart
from a vacuum. All electromagnetic waves
travel at the same speed in a vacuum. This
is referred to as the speed of light, c, and it
is equal to 299 792 458 m s–1. For most
purposes, the speed of light is approximated to 3 × 108 m s–1.
While electromagnetic waves may travel
at the same speed, they vary widely in
wavelength and frequency. The visible
spectrum — the range of electromagnetic
radiation to which human eyes respond —
makes up only a very small section of the
electromagnetic spectrum.
550
600
1m
1 km
650
104
105
106
700
107
FIGURE 10.3 The Pleiades open cluster in the constellation
Taurus. All stars are incandescent sources of light.
AS A MATTER OF FACT
Generally, solid objects start to emit visible light when their temperature reaches about 525 oC (called the Draper
point). At temperatures lower than this, electromagnetic radiation is still being released by warm objects but,
because the light waves produced are in the infra-red part of the spectrum, we are unable to see the objects.
Non-luminous (or illuminated) objects don’t produce
their own light; instead, they are reflectors of light produced by a luminous source. A non-luminous body can
only act as a light source when there is a luminous body
present. The source of the moonlight that allows us to
see in the night is actually light from the Sun reflected
from the surface of the Moon. When we wish to read a
book in a dark room, we turn on a lamp whose light then
reflects from the white pages of the book, allowing us to
see it.
FIGURE 10.4 When liquid Luminol comes into
contact with the iron in blood haemoglobin in
the presence of ultraviolet light, the blood is
luminous as a result of the chemical reaction
rather than because of incandescence.
FIGURE 10.5 The Moon is a non-luminous object; it
reflects light from the Sun.
FIGURE 10.6 An image seen through a night-vision
device.
10.2.3 The ray model of light
When encountering a surface, light can exhibit a number of different behaviours. It may be reflected from the surface, it may
be absorbed by the material, it may pass through the material,
or it may exhibit a combination of these behaviours. What light
does when it enters a new medium depends upon the nature of
the material and the condition of the interface between the
media.
All light travels in straight lines. As a result, we can only see an
object if the light from that object can travel directly to our eye.
You can easily see the book that is right in front of you on the
desk, but you won’t be able to see the people sitting behind you.
You may not even be able to see the people sitting beside you
FIGURE 10.7 Rays are straight lines
indicating light propagating from a
light source. The further apart they
are, the dimmer the light.
Light
Bulb
TOPIC 10 Ray model of light 217
FIGURE 10.8 We can see an object only if light
from that object is able to travel to our eyes. This
person cannot see the yellow objects.
FIGURE 10.9 A ray diagram of a light bulb.
FIGURE 10.10 The normal is an imaginary
line drawn perpendicular to a surface where a
light ray is incident upon it.
Incident
light ray
Surface
clearly. However, while light may only move in a straight
line, we are able to manipulate its path and see what was
previously hidden by exploiting light’s ability to be
reflected and refracted.
As light travels in straight lines, it is often convenient to
represent the paths taken by light as rays when drawing
Normal
diagrams. This will be particularly useful in our next sections. Remember that the ray is only a representation.
Although the diagram in figure 10.9 shows nine rays leaving the surface of the light bulb, it should be
understood that this is only to imply the spreading nature of the light. We could as easily have drawn ten or
a hundred rays in a similar way.
It is also handy at this point to define the normal to a surface. The normal is an imaginary line drawn
perpendicular to a surface at the point where a light ray is incident upon it. We will be using the concept of
the normal a great deal over the rest of the chapter.
10.2.4 Transmission of light through a medium
Light that strikes the surface of a material and is not reflected may pass through it. We are able to see
through glass windows because light rays travelling from outside objects are able to pass through easily
to our eyes. However, we cannot see what lies on the other side of a brick wall because light rays are
218 Jacaranda Physics 11
unable to pass through it. The ability of a material to allow light to pass through it is referred to as its
optical transmissivity. This is dependent upon the arrangement, type and size of atoms that the material is
made from.
A material through which light rays are able to travel without distortion of their relative pathways is said
to be transparent. Glass, Perspex and cling wrap are all examples of transparent materials. We are able to
see and identify objects through them without losing clarity.
Some materials, such as frosted glass and thin rice paper, allow us to see that there is an object on the
other side of them, but light rays cannot pass through them easily enough to see a clear image of the object.
Such materials are said to be translucent. These translucent materials allow light rays to pass through
them, but irregularities in their structure cause the rays to be scattered as they do so. As these scattered rays
emerge, they are able to give a vague impression of the object but not a clear picture.
Opaque materials are those which light is unable to pass through at all. Wood, brick, concrete and the
human body itself are all opaque. (Some parts of the human body, such as skin and nails, are translucent
when they are separate from the rest of the body.)
FIGURE 10.11 A teddy bear as seen through materials that are (a) transparent, (b) translucent and (c) opaque
(a)
(b)
(c)
Upon striking an opaque material, light rays may be reflected from it or, in some cases, they may be
absorbed completely. A black object with a rough surface will tend to absorb light, allowing almost none of
it to be reflected back to the observer’s eye. The absorbed light energy is usually converted into heat energy.
Light rays striking an opaque material cannot be transmitted through it.
WORKING SCIENTIFICALLY 10.1
Baking paper is a translucent material that allows a limited quantity of light from a light source to pass through
it. If enough layers of baking paper are placed over one another, they act as an opaque medium. Use a light
meter to measure the amount of light that is transmitted from a light bulb or other consistent light source
through a single layer of baking paper. Investigate the relationship between the number of layers and the
amount of light transmitted.
TOPIC 10 Ray model of light 219
10.2.5 Light intensity
The luminous intensity, I, of light is a quantitative measure of the effective brightness of a light source and
takes into account the amount of light energy produced by the light source each second and the area over
which that light energy is distributed.
The amount of light energy in joules produced by a light source each second is measured by its luminosity, L. The luminosity of a light source (also referred to as its luminous power) is measured in joules
per second or watts (W).
Like mechanical waves such as sound, light waves travel
FIGURE 10.12 Light obeys an inverse-square
outwards from their source in three dimensions, forming a
relationship with distance.
spherical wavefront, with the light energy distributed each
The Inverse-Square Relationship for Light
second over a larger and larger area. The area of this spherical
Sphere, distance r from source
3r
wavefront increases with distance from the source:
2r
r
Asphere = 4πd2
A
As the intensity of the light can be described as the
amount of energy distributed over each square metre of the
wavefront each second, we can then describe the intensity
in terms of the luminosity of the light source and the
observer’s distance away from it:
L
I=
4 πd2
with intensity having the units W m–2.
r
A
Light source
A
A
AA
A
A
A A
A A
A
A
Intensity ∝ 1/r2
At a distance 2r from the source the radiation is spread
over four times the area so is only 1/4 the intensity that
it is a distance r.
10.2 SAMPLE PROBLEM 1
A white sheet of paper is held 30 cm from a 60 W reading light. By what factor is the intensity of the
light incident on the paper reduced if the paper is moved 50 cm further away from the reading light?
SOLUTION:
First, the intensity of the light at the two positions can be described by:
L
I1 =
4 π (d1)2
and
L
I2 =
4 π (d2)2
As the luminous power of the source is the same in both cases, the two equations can be combined
by substituting for L and cancelling common terms to get the relationship:
I1 (d1) 2 = I2 (d2) 2
and so,
I1
d 2
= 2
I2 (d1)
Substituting in values, we find
I1
0.8 m 2
=
I2 (0.3 m)
I1
= 7.1
I2
Therefore, by moving the paper 50 cm further away, the intensity of light on the paper has been
reduced by a factor of 7.1.
220 Jacaranda Physics 11
WORKING SCIENTIFICALLY 10.2
Only a fraction of the energy produced by a light bulb is in the form of light energy; most of the energy is lost as
heat energy. Devise a method to determine exactly what proportion of light energy is produced by a light bulb
and then use this method to investigate the relationship (if any) between the wattage of a light bulb and the
proportion of light energy it produces.
10.2 Exercise 1
1 Which of the graphs in figure 10.13 best describes
FIGURE 10.13
the relation between luminous intensity and distance
from the light source?
A
B
2 Which of the graphs in figure 10.13 best describes
the relation between luminous intensity and the
power of the light source?
3 Which of the following are not luminous bodies in
the visible spectrum:
(a) a star
(b) the Moon
C
D
(c) a candle
(d) an incandescent light globe
(e) an LED light bulb
(f) coals that are hot but not glowing?
4 Place these forms of electromagnetic radiation in
order of increasing wavelength: ultraviolet, infrared,
microwaves, gamma rays, visible light, X-rays
5 Which of the following materials can be described
as translucent:
(a) frosted glass
(b) crystal glass
(c) cardboard
(d) steel?
6 The Earth is located 150 million kilometres from the Sun. How long does light from the Sun take to
reach the Earth?
7 A light is placed 40 cm from a screen. If the luminous intensity of light falling on the screen is 25 W m–2,
what will be the luminous intensity of light on the screen if the light source is placed 70 cm from the
screen?
8 A 20 W light bulb illuminates the page of a book with a luminous intensity of 5 W m–2. What will be the
luminous intensity incident on the page if the 20 W bulb is replaced with a 50 W bulb?
9 Two light sources, A and B, are placed either side of a white screen. When source A is placed 5 m from
the screen, it provides the same luminous intensity as source
B does on its side of the screen. If source B is ten times
stronger as a light source than source A, how far away from
the screen has source B been placed?
FIGURE 10.14
10 The Sun has a luminosity of 3.846 × 1026 W and is located 150
million km from the Earth.
(a) Calculate the intensity of the Sun’s light on the Earth.
(b) Assuming that the Earth is a sphere with a radius of 6370 km,
calculate the amount of solar energy that falls on the surface
of the Earth each second. (Hint: remember that only half of the
Earth’s face is illuminated by the Sun at any one time.)
11 Draw light rays to represent light shining from the reading lamp
shown in figure 10.14.
12 Draw a diagram showing the normal to the surface at the point
where a light ray is incident upon it.
TOPIC 10 Ray model of light 221
10.3 Reflection
10.3.1 The Law of Reflection
Reflection is, essentially, the bouncing of light
from a surface. Light rays from luminous
objects enter our eyes directly but we see
non-luminous objects because some of the
light rays falling on them from luminous
sources are reflected into our eyes. We see the
Moon because light from the Sun strikes the
surface and bounces (reflects) off it. Some of
this reflected light reaches our eye and, so, we
see the Moon. Light may bounce from several things before it reaches our eyes. A
moonlit river is the result of light from the
Sun bouncing off the Moon, and this moonlight then reflecting from the river to us.
The degree to which the light rays falling
on a non-luminous object are reflected from
it depends upon the nature of the object’s
surface. In some cases, the surface is such
that very little of the light falling upon it is
reflected at all — instead, most of the energy
from the light rays is absorbed by the surface.
For example, when light from the Sun
falls on a black bitumen road, most of the
light is absorbed, so very little of the Sun’s
light is reflected from it to our eyes. As a
result, we perceive the bitumen road to be
very dark. Conversely, almost all the sunlight
falling on packed snow is reflected, so it
appears very bright to our eyes.
Under ideal conditions, a surface may be
so smooth as to reflect nearly all the light
incident upon it, allowing a clear image to be
formed.
The geometry of perfect reflection is summarised in the Law of Reflection, which
states that the angle between the incident
light ray and the normal where it strikes the
surface will be equal to the angle between
the normal and the reflected ray.
We can express this mathematically as
i = i′
where i is the incident angle formed between
the incoming ray and the normal, and i′ is the
angle of reflection.
222 Jacaranda Physics 11
FIGURE 10.15 Light rays from a luminous object hitting a
book and reflecting into an eye.
FIGURE 10.16 Diagram showing incident and reflected
rays for a plane surface.
normal
reflected
ray
incident
ray
angle of
incidence
angle of
reflection
i
i′
mirror
normal
reflected ray
mirror
incident ray
10.3.2 Types of reflection
Reflection of light rays from a non-luminous surface
FIGURE 10.17 Regular reflection occurs when
can be described as being either specular or diffuse.
parallel incident rays are reflected from a surface in
Specular reflection (also referred to as regular
such a way that the reflected rays are also parallel.
reflection) is observed from surfaces that are very
Incident rays
Reflected rays
smooth or highly polished, such as still water and
good-quality mirrors. When parallel rays strike
these surfaces, each ray is reflected evenly from the
surface so the reflected rays are also parallel to each
Surface
other. The more polished and even the surface, the
less we see the surface itself and the more we see
FIGURE 10.18 Aurora Borealis reflected in a
the reflection of other objects in it instead.
perfectly calm fjord on a cold winter night.
Surfaces that are irregular will not reflect parallel
incident rays uniformly but will scatter them. We call
this type of reflection diffuse (or irregular) reflection.
When each incident ray strikes the surface, it
obeys the Law of Reflection. However, as the section
of surface that each ray strikes is angled differently
due to the surface irregularity, the normal for each
point of incidence will not be parallel to the normal
for the next section. As the angle of incidence for
each ray is different, the reflected angles for the rays
will also differ, causing the reflected light to be scattered. Only some of these reflected rays will reach
our eyes. Thus, while a very white piece of ‘smooth’ paper will reflect enough diffuse rays to our eyes to
appear bright and light, the irregularity of those rays prevents the formation of a coherent reflected image in it.
10.3.3 Images formed by plane mirrors
FIGURE 10.19 The image of your face in the
When you look in the mirror, what you see is a reflected
mirror is an optical illusion caused by the
reflection of light in the mirror. It is called a
image of your face. Assuming that the mirror you are
virtual image.
looking at is a plane (flat) mirror, you can see that the
image is about the same size as you would expect your
head to be when viewed at that distance, and it is the right
way up. You might also notice that the image appears to
be behind the mirror at the same distance behind it as you
are in front of it. You know that there is not really an
image behind the mirror. That space is probably in another
room or outside. The image of your face in the mirror is
an optical illusion caused by the reflection of light in the
mirror. This image is called a virtual image.
The ray model helps us to understand how this image
forms. In figure 10.20 you can see the object (your head)
and the mirror from a view to one side. Your head is
non-luminous, but because you are in a lit room, light
striking your head is diffused in all directions. Consider
light striking the top of your head. Some of this light
reflects in the direction of the mirror. We can choose to investigate the behaviour of any of the rays that hit
the mirror, but let’s start with the ray that passes horizontally to the mirror (ray 1). It will reflect with i = i′
TOPIC 10 Ray model of light 223
so that the reflected ray retraces the path of
FIGURE 10.20 Locating the image in a plane mirror.
the incident ray. Since this ray returns to the
top of your head, it never actually enters
your eye, so it does not contribute to the
I
1
image formed by your eye.
o
Now consider what happens to a ray of
2
light that passes from the top of your head
3
to the mirror and reflects back to your eye
(ray 2 ). Again, we know its path because
i = i′. This ray helps to form the image that
your eye sees.
Consider another ray that travels from the top of your head to the mirror and reflects back to touch your
chin (ray 3). Again, this ray does not enter your eye.
What we can see is that all three rays can be traced back to a single point behind the mirror. This point,
labelled I, is exactly where we see the image of the top of our head in the mirror. There is nothing special
about the three rays chosen. Draw any other ray and trace back its reflected ray, and we see that it too
appears to come from this point.
Only one ray that we drew enters the eye. How can the eye form an image of the top of the head from a
single ray? A ray represents an infinitesimally small beam of light. Many rays of light enter the pupil of the
eye, all from slightly different angles, so the eye can interpret them as diverging from a point behind the
mirror.
When drawing diagrams such as this, it
FIGURE 10.21 Light diverging from a virtual image to your
eye.
is much easier to use rays of light that are
well spread out, even if they do not enter
I
the eye. It does not make any difference to
the result.
We have now located the position of
the image of the top of your head using the
technique of ray tracing. We can do the
same for the chin. See figure 10.22. What
we find is that the image is the same size as
the object, it is upright and appears to be at
the same distance behind the mirror as the
FIGURE 10.22 Locating the image of your chin.
object is in front of the mirror. It is also a
virtual image, because the light only appears
to come from the image. In reality, the light
from your head does not pass through the
image at all.
I
Next we will investigate the formation of
o
other types of images. Real images are
actually formed by the light rays. These are
essential in the eye and in cameras, both of
which have sensors that respond to the light
of the image. We can only see virtual images
because our eyes make real images of the light appearing to come from virtual images.
An interesting fact about plane mirrors is that the image is laterally inverted. This means that the lefthand side of the object is the left-hand side of the image, but the image is facing the object. So if you wear
a watch on your left hand (the object), the image will have the watch on its right hand. This is simply
explained by drawing a ray diagram as seen from above the situation.
224 Jacaranda Physics 11
Images are not always the same size as their objects. The effect of an optical device on the size of the
image is indicated by the magnification:
M=
H1
H0
where H1 is the height of the image and H0 is the height of the object. As the image in a plane mirror is the
same height as the object, the magnification is 1. In a device such as the bottom of a spoon, where the
height of the image is smaller than the height of the object, then the magnification is between 0 and 1. This
is known as a diminished image. When the magnification is greater than 1, the image is said to be enlarged.
If you look at the reflection of your eye in the concave (curved inwards) side of a polished spoon, with
your eye very close to the spoon, you may see an enlarged image of your eye.
10.3 SAMPLE PROBLEM 1
Joan is 160 cm tall and her eyes are located a distance 8 cm from the top of her head. What will be the
shortest length mirror that she can purchase in order to see a full-length image of herself?
SOLUTION:
FIGURE 10.23
8 cm
A
AA
A′
E
ZZ
152 cm
Z
d
Z′
d
M
I
We can represent Joan’s length by a line AZ with the point E marking the position of her eyes as
shown in the figure above.
We know that her image will be located the same distance from the mirror M as she herself is so. If
we say that she is a distance d from the mirror when she sees a complete image of herself, then we
know her image must be located on the line I.
Light rays leaving A must reflect back to point E, as must those leaving Z if Joan is to see an image
of them in the mirror. Joan’s eye will see the image of A at A′ and of Z at Z′. The rays reflected from A
and Z back to E will appear to come from A′ and Z′, and the reflection points will be at AA and ZZ. The
distance between AA and ZZ is the minimum length of the mirror that Joan needs to buy. Using geometry, we can calculate this distance.
We can see that ZZ will be located at a height halfway between E and Z (which will be 76 cm from
the ground), while AA will be located halfway between A and E (a point 4 cm below Joan’s head
height). This means that AA and ZZ are 80 cm apart.
Hence, Joan will need to buy a mirror that is at least 80 cm long if she is to see a full length-image
of herself in it.
TOPIC 10 Ray model of light 225
10.3 Exercise 1
1 A ray of light strikes the surface of a plane mirror so that the angle between it and the mirror is 35 °.
(a) Determine:
(i) the angle of incidence
(ii) the angle of reflection.
(b) What do the incident ray, the normal and the reflected ray all
FIGURE 10.24
have in common (other than being straight lines)?
2 A man who is 170 cm tall stands 50 cm in front of a plane
mirror mounted on a wall. What is the shortest mirror that
45°
could be used if he is to see his entire body reflected? Assume
that his eyes are 5 cm down from the top of his head.
3 Explain why the lettering on the front of emergency vehicles is
Object
written back-to-front.
4 Explain why you can see your reflection in a highly polished sheet
of silver metal but not in a sheet of paper.
5 An object is placed between two mirrors that form a 45 o angle
between them. Copy the diagram in figure 10.24 and use ray
tracing to locate all the images formed in the mirrors.
10.4 Curved mirrors
10.4.1 Concave and convex m
­ irrors
Sometimes mirrors are used for purposes that require them to be curved rather than flat plane mirrors.
Curved mirrors can be either concave (with the polished surface on the interior curve like inside the bowl of
a spoon, as shown in figure 10.25a) or convex (where the polished surface is on the outside curve, as shown
in figure 10.25b).
FIGURE 10.25 Reflections in concave and convex surfaces.
(a)
(b)
Most curved mirrors are shaped as sections from a sphere or an ovoid; for the moment, we will concentrate on spherical mirrors only.
To better understand how curved mirrors reflect incident rays, consider a concave mirror to perform similarly to a series of plane mirrors arranged in a curve, as shown in figure 10.26.
Parallel incident rays striking these individual plane mirrors will be each reflected according to the angle at
which they strike the mirror. For each mirror, the angle at which the reflected ray leaves the mirror surface
will be equal to the incident angle. As each mirror is arranged at a different angle to the one beside it, the
reflected rays will not be parallel; rather, they meet at a common point referred to as the focus, F, of the
mirror.
226 Jacaranda Physics 11
Because concave mirrors cause parallel rays
FIGURE 10.26
FIGURE 10.27
to come together to a point (or converge), they
are also referred to as converging mirrors
(see figure 10.27).
Convex mirrors, on the other hand, cause
parallel rays striking the surface to be spread
out on reflection, and so they are called
diverging mirrors (see figure 10.28).
The spreading rays reflected from a convex
incident rays
mirror appear to originate at a common point inside the convex mirror.
reflected rays
As for the concave mirror, this common point is referred to as the
FIGURE 10.28
focus. The main difference between the foci of the two mirror forms
is that reflected light rays actually intersect at the focus of the concave mirror while, for the convex mirror, the focus is virtual and the
reflected light rays do not intersect there.
F
10.4.2 Mirror terminology
The geometry of concave (and convex) mirrors is critical to the way
F
in which they focus incoming light and form images, so it is useful to
describe the main features of a spherical mirror as follows:
•• The centre of curvature (C) is the geometric centre of the sphere
of which the curved mirror is a section.
•• The optical centre (O) is the centre of the curved mirror’s face.
•• The radius of curvature (R) is the radius of this sphere; this will
be the distance between the centre of curvature and the geometric
centre of the mirror.
incident rays
•• The principal focus (F) is the point at which the reflected rays conreflected rays
verge when the incident rays are parallel to the principal axis (condirection from which reflected
cave mirror) or the point from which diverging reflected rays
rays appear to come
appear to originate (convex mirror).
•• The principal axis is the line upon which the centre of curvature, the principal focus and the optical
centre lie.
•• The focal length (f) is the distance between the principal focus and the optical centre. For a spherical
mirror, the focal length is half the radius of curvature.
FIGURE 10.29 The geometric features of a spherical mirror.
Concave (converging) mirror
principal axis
F
C
O
Convex (diverging) mirror
principal axis
F
O
C
R
f
R
f
TOPIC 10 Ray model of light 227
FIGURE 10.30 The focal plane of a spherical mirror.
Focal plane
Focal plane
F
O
O
F
Parallel rays incident on a spherical mirror but which do not approach parallel to the principal axis will still
be focused at a point. However, rather than intersecting at the focal point of the mirror, the reflected rays
intersect at a position on the focal plane.
10.4.3 Ray tracing
When an object is placed in front of a curved mirror, an image may be formed. This image may be real or
virtual depending upon both the distance the object is placed in front of the mirror and the type of mirror
being used.
Scale diagrams incorporating the paths taken by light rays can be used to determine the characteristics of
the image formed by each type of mirror at different object distances. This process is referred to as optical
ray tracing.
While an infinite number of light rays could be drawn travelling from an object to the mirror, the paths of
four rays, in particular, are the most easily traced:
1. Any ray that travels parallel to the principal axis from the object to the mirror will be reflected so that it
passes through the mirror’s focus.
2. A ray that passes through the focus as it travels from the object to the mirror will be reflected so that it
travels parallel to the principal axis.
3. Rays that travel through the centre of curvature as they travel
FIGURE 10.31
from the object to the mirror are reflected back along their
original path.
4. A ray that travels from the object to the optical centre is
reflected so that the angle made between the reflected ray and
the principal axis is equal to the incident angle.
As an example, let’s look at how ray tracing can be used
to find the type, height and orientation of an image formed
by a converging mirror of an object located between the
focus (F) and the centre of curvature (C). In this case, let us
C
assume that the mirror has a focal length of 3 cm.
Step 1. First, a horizontal line is drawn that represents the
principal axis. A point is marked on the principal axis to represent
the centre of curvature (C) of the mirror. As we know the focal
length of the mirror is 3 cm, the radius of curvature of the mirror
will be twice that — that is, 6 cm. Using a compass, we draw a
part circle with a radius of 6 cm and centred on point C, as
shown in figure 10.31.
228 Jacaranda Physics 11
Step 2. The focus (F) is marked on the principal axis 3 cm away from the centre
FIGURE 10.32
of the mirror. The object (represented by a thick arrow, OP) is drawn with its
base on the principal axis, as shown in figure 10.32.
Step 3. Two rays are drawn from the top of the object at P, as shown in
figure 10.33:
•• Ray 1 leaves the head of the object parallel to the principal axis, is reflected
P
by the mirror and passes back through the focus.
•• Ray 2 passes through the focus, is reflected by the mirror, and then travels
O F
C
back parallel to the principal axis.
Where these rays intersect, the image of P (P)′ will be formed. Note that while
two, three or even all four of the main ray paths may be employed to locate the
image position, the clarity of the diagram drawn can be lost.
Step 4. As OP was placed perpendicularly to the principal axis, the image of O
(O)′ will be located on the principal axis such that O′P′ is also perpendicular, as
shown in figure 10.34.
The image O′P′ formed is a real image because light rays actually pass through it. If a screen was placed
at that location, the clear image would be seen upon it.
The image is upside-down compared to the object, so we say that it has been inverted.
The image can also be described as enlarged or dilated because the height of O′P′ is greater than that of
OP. The image is located at a distance greater than 2f.
In a similar way, scaled ray diagrams can be drawn to determine the location (di), height (hi) and the
nature of the image formed by an object placed at a distance do from a converging mirror and having a
height of ho. The formation of these images is summarised in Table 10.1
FIGURE 10.33
FIGURE 10.34
P
P
O′
C
P′
ray 1
O
C
F
ray 2
ray 1
P′
O
F
ray 2
TABLE 10.1 Images formed by a converging (concave) mirror.
Object
Ray diagram
Image
position
position
Inverted or
upright?
Size
Real or
virtual
do > 2f
2f > di > f
inverted
hi < ho
real
P
Q′
Q
C
P′
F
O
(continued)
TOPIC 10 Ray model of light 229
TABLE 10.1 Images formed by a converging (concave) mirror.
Object
Ray diagram
position
Image
position
Inverted or
upright?
Real or
virtual
Size
P
Q
Q′ C
O
F
do = 2f
di = 2f
inverted
hi = ho
real
2f > do > f
di > 2f
inverted
hi > ho
real
do = f
di = ∞
(no image formed)
do < f
di > − 2f
upright
P′
P
Q′ C
F
Q
O
P′
P
Q
C
F
O
P′
P
F Q
O
hi > ho
virtual
Q′
While concave mirrors can form both real and virtual images
depending upon how close the object is to the mirror, convex mirrors
FIGURE 10.35
can only form virtual images.
Ray diagrams for convex mirrors are drawn in much the same
P
way as for concave mirrors. However, as convex mirrors have no
true focus, the reflected paths taken by rays travelling from the
Q
object to the mirror will be related to the virtual focus, which
appears to lie inside the mirror.
When an object PQ is placed in front of the diverging mirror,
as shown in figure 10.35, we consider the paths taken by two rays
FIGURE 10.36
leaving P and travelling to the mirror, as shown in figure 10.36:
ray 1
•• Ray 1 leaves P and travels parallel to the principal axis. On
reaching the mirror, the ray is reflected so it appears to travel
P
ray 2
through the focus.
•• Ray 2 travels in a straight line directed from P to the focus; on
Q
striking the mirror, the reflected ray travels parallel to the
principal axis.
230 Jacaranda Physics 11
C
F
F
C
The intersection of the continuing lines of ray 1 and ray 2 marks
FIGURE 10.37
the location of the image of P (P′). As PQ is perpendicular to the
ray 1
principal axis, Q′ can be located directly underneath P′, as shown
in figure 10.37.
P
The virtual image P′Q′ formed in this case is smaller than the ray 2
P′
object PQ and is positioned closer to mirror.
Q
Q′ F
C
All images formed by convex mirrors are:
•• virtual
•• upright
•• reduced in height.
FIGURE 10.38 Convex mirrors are often used to
10.4.4 The mirror equations
help improve vision around corners.
While ray tracing provides qualitative information
about the position, size and nature of the images
formed by concave and convex mirrors, more precise
numerical information about the position and size of
the images can be obtained by use of the mirror equation and the magnification equation.
The mirror equation relates the distances of the
object (do) and the image (di) from the mirror to the
mirror’s focal length f :
1
1
1
=
+
f
do
di
In general:
•• do has a positive value
•• for a concave mirror, f is a positive value (as it has a true focus where light rays intersect) while a
convex mirror has a negative value for the focal length
•• for a virtual image, di will have a negative value. This is the case when an object is placed within the
focus of a concave mirror, and for all images formed by a convex mirror.
The magnification M describes the height of the image (hi) relative to the height of the object (ho):
h
M= i
ho
The magnification can also be determined from the distances of the object and the image from the mirror:
d
M=− i
do
For a concave mirror, all real images will be inverted and hi has a negative value, while virtual images
are upright and hi has a positive value.
By combining the two equations, we find a third form of the magnification equation:
hi
d
=− i
do
ho
10.4 SAMPLE PROBLEM 1
A 4 cm high object is placed 10 cm in front of a converging mirror having a radius of curvature of 16 cm.
(a) Calculate where the image forms relative to the mirror.
(b) Calculate the height of the image.
(c) Determine whether the image is
(i) real or virtual,
(ii) upright or inverted, and
(iii) enlarged or reduced.
TOPIC 10 Ray model of light 231
SOLUTION:
(a) As the focal length of a spherical mirror is equal to half the radius of curvature,
16 cm
= 8 cm
f=
2
As the mirror is a converging mirror, the focal length will have a positive value.
Substituting values into the mirror equation:
1
1
1
=
+
f
do
di
1
1
1
=
+
8 cm
10 cm
di
1
1
1
=
−
8 cm
di
10 cm
1
1
=
di
40 cm
di = 40 cm
The image will be located 40 cm from the mirror.
(b) Using the magnification equations:
hi
d
=− i
do
ho
hi
40 cm
=−
4 cm
10 cm
hi = –4 × 4 cm
= –16 cm
The image has a height of 16 cm.
(c) (i) As di > 0, a real image has formed
(ii) As hi < 0, the image is inverted
(iii) The image is 16 cm high while the object is only 4 cm high, therefore the image is enlarged.
10.4 SAMPLE PROBLEM 2
A 5 cm object is placed 12 cm from a diverging mirror that has a focal length of 10 cm. Determine the
location, height and nature of the image formed.
SOLUTION:
As the mirror is convex (diverging), the focal length will have a negative value: f = –10 cm.
Substituting values:
1
1
1
=
+
12 cm
−10 cm
di
1
1
1
=−
−
12 cm
di
10 cm
1
11
= − cm
di
60
60
di = − cm
11
= −5.4 cm
hi
d
=− i
do
ho
hi
−5.4 cm
=−
12 cm
5 cm
232 Jacaranda Physics 11
hi = 2.25 cm
h
2.25 cm
M= i =
= 0.45
ho
5 cm
Therefore, the image is 2.25 cm high and it appears to form 5.4 cm inside the mirror. The image is
virtual (as di < 0), upright (as hi > 0) and reduced in size (as M < 1).
WORKING SCIENTIFICALLY 10.3
Archimedes is famously credited with using a series of spherical concave mirrors to set enemy ships alight.
Some translations of the story give the distance from the mirrors to the ships as being equivalent to 1.5 km.
Assess the feasibility of this story by considering the distance over which such a feat might be possible, the
diameter of the mirrors, the number of mirrors needed, their radius of curvature, and the temperature achievable
by focusing the rays of the Sun.
10.4 Exercise 1
1 Dentists often place a small mirror inside the patient’s mouth to examine their teeth. Is the mirror used
more likely to be concave, convex or flat?
2 A lit match is placed 5 cm from a converging mirror that has a radius of curvature of 10 cm. Which best
describes the image:
(a) real, reduced and inverted
(b) real, enlarged and inverted
(c) virtual, enlarged and upright
(d) no image is formed?
3 Which of the following statements is true:
(a) concave mirrors can only form real images
(b) convex mirrors can only form virtual images
(c) the image formed by a plane mirror is always the same size as the object
(d) the image formed in a plane mirror always appears to be the same distance from the mirror as the object
(e) concave mirrors can be used as magnifying makeup mirrors
(f) an object placed at the focus of a converging mirror reflects rays parallel to the principal axis
(g) the Law of Reflection is only true for plane mirrors and does not apply to spherical mirrors?
4 A 3 cm high object is placed 6 cm from a diverging mirror with a focal length of 4 cm. How high is the
image formed as a result?
5 An object placed 6 cm from a converging mirror forms a real image 10 cm from the surface of the mirror.
What is the mirror’s focal length?
6 How far from a concave spherical mirror with a focal length of 12 cm must an object be placed to
produce a virtual image that is 3 times larger than the object?
7 Use ray tracing to show the approximate location and nature of the image formed when an object is
placed 4 cm in front of a spherical converging mirror with a focal length of 3 cm.
8 Use ray tracing to show the approximate location and nature of the image formed when a 6 cm high
object is placed in front of a spherical diverging mirror with a focal length of 4 cm.
RESOURCES
Complete this digital doc: Model of a concave mirror
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Concave mirror applet
Convex lens applet
TOPIC 10 Ray model of light 233
10.5 Refraction
10.5.1 The speed of light
Visible light travels in a vacuum at the same
speed as all other electromagnetic radiation —
3 × 108 m s–1. When it encounters any other
medium, it will slow down. The degree to
which the speed of light is slowed when it
moves through a material is described by the
absolute refractive index (n) of the material.
This value is the ratio of the speed of light in a
vacuum (c) compared to its speed in the
medium (v):
c
n=
v
Table 10.2 shows the absolute refractive
indices of some common media.
TABLE 10.2 Absolute refractive indices.
Material
Index of refraction
Vacuum
1.00
Air*
1.00
Water
1.33
Quartz
1.46
Car headlight glass
1.48
Perspex (average)
1.50
Window glass
1.51
Crystal wineglass (24% lead)
1.54
Diamond
2.42
* The slowing of light in air is fairly small and, for most cases,
can be assumed to be negligible.
10.5 SAMPLE PROBLEM 1
Light travels at a speed of 2.26 × 108 m s–1 in water. Calculate water’s absolute refractive index.
SOLUTION:
3 × 108 m s−1
2.26 × 108 m s−1
= 1.33
The refractive index of water is 1.33. This means that light travels 1.33 times faster in air than it
does in water.
n=
10.5.2 The bending of light
Refraction refers to the bending of light that occurs when light travels through transparent media that have
different refractive indices. The reason that the light bends is connected to the fact that light travels at different speeds in different media.
We’re going to use an analogy at this point to help us understand how a changing speed leads light to
bend when travelling through different media.
You may have noticed that a four-wheel drive travels faster over packed wet sand on the beach than it
does over dry loose sand. In this way, the four-wheel drive is much like light in that it will travel more
slowly through some media than others. Now, let’s say that the four-wheel drive is travelling along a section of wet sand when it comes to a section of dry sand. It is headed towards the demarcation line between
the two types of sand at an angle i as shown in Figure 10.39a.
The first tyre to hit the dry sand will be the right front tyre in our diagram. As soon as it enters the dry
sand region, it will start turning more slowly than the other wheels (figure 10.39b). This has the effect of
causing the front of the car to be dragged off course, and it will veer to the right as it enters the dry sand
(figure 10.39c). As a result, the course of the car has been altered.
Light entering a new medium will behave similarly to the four-wheel drive on the beach. If we could
look at the light waves as they strike the interface between media, we would see that they too are diverted
from their course.
234 Jacaranda Physics 11
FIGURE 10.39 A four-wheel drive entering an area of dry sand.
(b)
(a)
i
(c)
Wet sand
Wet sand
Wet sand
Dry sand
Dry sand
Dry sand
Figure 10.40 shows a light wave entering a medium in
FIGURE 10.40 The refraction of light at a
which it travels more slowly. The line AB represents a waveboundary between different media.
front approaching the interface between air and glass. The
section of the wavefront at A strikes the boundary before
that — at point B. On entering the glass, the light waves at
A will slow down while the rest of each wave continues to
travel through air at the original, faster speed. During the
B
time taken for the waves at A to travel to position C in the
Air
D
A
new medium, the waves at the other end of the wavefront
Glass
C
have travelled a larger distance from B to D. As a result, the
wavefront changes direction as it crosses the interface.
The extent to which light is bent when it enters a second
medium depends upon the speed of light in the individual
media. As you will recall, the speed of light in a medium
can be related by the absolute refractive index (n) of that material.
When light strikes an interface between media at an angle i (which is the angle between the incident ray
and the normal), it will be refracted so that the transmitted light will travel at the refracted angle r (the
angle between the refracted light and the normal). If light travels from a lower refractive index medium to
a medium with a higher refractive index, it will bend towards the normal — that is, if n2 > n1 then r < i.
Conversely, if the second medium has a refractive index that is lower than that of the first medium, then the
light will be bent away from the normal as it is transmitted — that is, r > i if n2 < n1.
FIGURE 10.41 Light entering a
medium with a higher refractive
index will be bent towards the
normal as it is transmitted.
FIGURE 10.42 Light entering a medium
with a lower refractive index will be bent
away from the normal as it is transmitted.
Normal
Normal
Incident ray
Incident ray
n1 < n2
n1 > n2
i
i
n1
n2
n1
n2
r
r
Refracted ray
Refracted ray
TOPIC 10 Ray model of light 235
10.5.3 Snell’s Law
The extent to which light is bent when it changes medium can be determined using Snell’s Law, which was
first formulated in approximately 1621 by the Dutch scientist Willebrord Snell:
n1 sin (i) = n2 sin (r)
where n1 is the refractive index of the incident medium, i is the incident angle, n2 is the refractive index of
the new medium, and r is the angle of refraction.
FIGURE 10.43 A graphical depiction of Snell’s Law for any two substances. Note that the light ray has no
arrow, because the relation is true for the ray travelling in either direction.
normal
medium 1
refractive index n1
i
boundary
medium 2
refractive index n2
r
n1 sin i = n2 sin r
10.5 SAMPLE PROBLEM 2
A ray of light strikes a glass block of refractive index 1.45 at an angle of incidence of 30°. What is the
angle of refraction?
SOLUTION:
,
1.0 × sin 30° = 1.45 × sin θ glass (substitute values into Snell s Law)
sin 30°
sin θ glass =
(divide both sides by 1.45, the refractive index of glass)
1.45
= 0.3448 (calculate value of expression)
θ glass = 20.17° (use inverse sine to find the angle whose sine is 0.3448)
θ glass = 20° (round off to two significant figures)
10.5 Exercise 1
Light travels at a speed of 2.1 × 108 m s–1 through medium X. What is this medium’s refractive index?
What will be the frequency of violet light (λ = 420 nm) as it passes through window glass?
Calculate the speed at which light travels through diamond.
How many times faster does light travel through glass than it does diamond?
Which of these statements is true:
(a) light rays entering a new medium change frequency
(b) light rays travel through glass at a lower speed than they do through a vacuum
(c) light rays entering a medium with a higher refractive index will be bent towards the normal
(d) light rays directed at right angles to the boundary between two media are not refracted?
6 Light travelling from water into glass (nglass = 1.53) is refracted at an angle of 49°. At what angle was the
light incident upon the glass?
1
2
3
4
5
236 Jacaranda Physics 11
7 If a laser light is shone onto a pool of water at an incident angle of 15° to the normal, what will be its angle
of refraction?
8 A beam of light shines onto a glass slab (nglass = 1.51) that has a thickness of 4 cm. If the beam makes an
angle of 30° with the slab surface, how far horizontally will the beam exit the block from where it entered?
9 A ray of light enters a plastic block at an angle of incidence of 40°. The angle of refraction is 30°. What is the
refractive index of the plastic?
10 In a science fiction story, a transparent material called ‘slow glass’ can slow down light rays entering the
material so much that they can take years to emerge from the other side. What would the refractive index
of such a material be if light entering a 20 cm thick pane of the glass took one day to emerge from the
other side?
RESOURCES
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10.6 Lenses
10.6.1 Converging and diverging lenses
The word ‘lens’ is a familiar one to anyone who wears glasses or has ever used a microscope or telescope.
A lens describes any transparent optical object with a curved surface that refracts light as it transmits it,
allowing redirection.
To begin to understand a lens, we can start with a rectangular block of glass as in figure 10.44. Parallel rays
from the left pass through the block without a change in direction if they are normal to the block (blue lines).
Parallel rays that are not normal to the block are refracted when passing through the block, but emerge parallel on
the other side (green lines). This is essentially what happens with light passing through a pane of glass in a
window.
If the block is shaped so that its surface is a continuous curve in the arc of a circle (figure 10.45a), all
sets of parallel rays entering the lens converge on the other side.
FIGURE 10.44
Lenses come in a variety of different forms, but can be generally classified as being either converging or
diverging.
TOPIC 10 Ray model of light 237
FIGURE 10.45 Refraction of rays through (a) a convex and (b) a concave lens.
(a) Convex lens
(b) Concave lens
F
F
The lens form we have considered so far
FIGURE 10.46 The
FIGURE 10.47 The
is referred to as a converging lens. A
converging lens in its
diverging lens in its
converging lens causes parallel light rays
different forms: (a) bi-convex,
different forms:
passing through it to be refracted towards a
(b) plano-convex, (c) convex
(a) bi-concave,
meniscus.
(b) plano-concave,
single point. As for a converging mirror,
(c) diverging meniscus.
this intersection point is referred to as a
focus. The converging lens comes in several
forms: the bi-convex, which has convex surfaces on each side; the plano-convex, which
has a convex shape on one side but is flat on
the other; and the converging meniscus,
which is convex on one side but concave on
(b)
(a)
(c)
(b)
the other. Regardless of their variations in (a)
(c)
shape, all converging lenses are thicker in
the middle than at their edges.
A diverging lens causes parallel light rays to be spread further apart after being refracted. The diverging
rays appear to come from a focus on the opposite side of the lens. Diverging lenses are thicker at their edges
than in their centres and can have a variety of forms: the bi-concave, which has both of its faces concave;
the plano-concave, where one of the lens’s faces is flat while the other is concave; and the diverging
meniscus, which has both concave and convex faces.
10.6.2 Lens terminology
FIGURE 10.48 The centre of curvature for a
lens face.
Many of the terms we will encounter in our study of
lenses will be familiar from our earlier study of mirrors:
•• The optical centre (or pole) of a lens is the point in the
exact centre of the lens itself. Light rays that pass
through the optical centre of a lens will not be diverted,
but will continue undeflected.
•• The centre of curvature (C) for the face of a lens is
the centre of the circle, an arc of which corresponds to
the curve of the lens face. A flat face of a lens has a
centre of curvature located at infinity.
•• The radius of curvature (R) is the distance between
the centre of curvature and the surface of the lens.
R
Lens
238 Jacaranda Physics 11
Centre of
curvature
•• The principal axis is a line that can be drawn through the centres of curvature for both faces of a lens
and the optical centre.
•• The focus (F) is the point at which light rays entering the lens parallel to the principal axis converge (or,
in the case of a diverging lens, appear to originate) on exiting the lens. Because light can pass through
either side of a lens, there is one focus on each side.
•• The focal length (f) is the distance between the optical centre and the focus. The focal length of a lens
depends upon the curvature of the lens faces, the thickness of the lens and the material from which it is
made. In general, the greater the curvature of the lens face, the shorter the focal length.
•• The focal plane is a plane through the focus that is perpendicular to the principal axis. When rays that
are parallel to one another enter the lens at an angle to the principal axis, they will converge at some
point on the focal plane.
FIGURE 10.49 The lens’s radius of curvature
influences the location of the focus.
FIGURE 10.50 Parallel rays entering a lens at an angle
to the principal axis will converge on the focal plane.
(a)
F
F
(b)
F
Focal plane
10.6.3 Images formed by converging lenses
Light rays passing from an object through a converging lens can form images of that object. However, the
orientation, size and nature of that image depend on how far the object lies from the lens and the focusing
ability of the lens itself.
As in the earlier section on images formed by mirrors, ray tracing can be used to give a qualitative
impression of the size and location of an image formed by a lens, as well as the nature of the image.
Four main principles are observed when using ray tracing for converging lenses:
1. Incident rays that travel parallel to the principal axis when approaching the lens will be refracted to pass
through the focus on the other side.
2. Incident rays that pass directly through the focus on the side nearest to the object as they approach the
lens will be refracted to pass parallel to the principal axis on the other side.
3. Incident rays that pass through the optical centre (pole) of a thin lens and that are incident at small angles
to the principal axis continue to travel in the same direction.
4. Images form where rays converge.
By considering these principles, a ray diagram can be drawn for an object PQ placed outside the focal
length of a converging lens and the image’s relative size and location determined. It should be noted that,
while refraction of rays occurs at each boundary between the air and the lens, in reality, the lenses in this
text are considered to be very thin. As a result, by convention, the bending of the light rays within the lens
TOPIC 10 Ray model of light 239
is represented by a single refraction at the lens axis (a line that passes through the pole of the lens that is
perpendicular to the principal axis).
From figure 10.51, we see that the image P′Q′ is located at a position greater than 2f and that it is
enlarged and inverted. The image is formed on the opposite side of the lens and is described as real. This
means that, should a screen be placed at the image position, an image will form on that screen.
The image obtained depends on the placement of the object in relation to the focus. A range of these
applications is given in table 10.3.
FIGURE 10.51 The location of the image is determined according to the point where the three rays
cross. All the rays that pass through the lens pass through the image.
object
convex lens
3
1
image
P
Q’
F
Q
F
2
P’
ray 2
ray 3
ray 1
TABLE 10.3 Simple applications of convex lenses
Location of object
Uses
Description of image
Very large distance away
from lens
Objective lens of refracting
telescope
Real, inverted, diminished and located near the
opposite focus
Beyond twice the focal
length from lens
At twice the focal length
from lens
Between twice the focal
length from lens and the
focus
At the focus
Human eye; camera
Real, inverted, diminished and located on other
side between one and two focal lengths from lens
Real, inverted, same size and located two focal
lengths from lens
Real, inverted, magnified and located on other
side of lens beyond two focal lengths
Between focus and lens
Correction lens for terrestrial
telescope
Slide projector; objective lens of
microscope
Searchlight; eyepiece of refracting
telescope
No image. The emerging parallel rays do not meet.
Magnifying glass; eyepiece lens of
microscope; spectacles for
long-sightedness
Virtual, upright, magnified and located on same
side of the lens and further away
PHYSICS IN FOCUS
Flat lenses?
A lens works by changing the direction
of the light ray at the front surface and
then again at the back surface. The glass
in the middle is there to keep the two
surfaces apart. Augustin-Jean Fresnel
devised a way of making a lens without
the need for all the glass in the middle.
240 Jacaranda Physics 11
FIGURE 10.52 A side view of a convex Fresnel lens showing
how it is constructed.
The glass surface of the lens is a series of concentric rings. Each ring has the slope of the corresponding section of the full lens, but its base is flat. The slopes of the rings get flatter towards the centre.
This design substantially reduces the weight of the lens, so lenses of this type are used in lighthouses. Their relative thinness means they are also used where space is at a premium, such as in overhead projectors, and as a lens to be used with the ground-glass screens in camera viewfinders.
Flat lenses, or Fresnel lenses as they are called, are now attached to the rear windows of vans and
station wagons to assist the driver when reversing or parking.
10.6 SAMPLE PROBLEM 1
A convex lens has a focal length of 10 cm. A candle 10 cm tall is located 16 cm in front of the lens.
Use ray tracing to determine the location, size, orientation and type of image formed.
SOLUTION:
Draw the principal axis, the focal points, the object and three rays, one passing through the centre of
the lens without deviation, one parallel to the principal axis and one passing through the focus.
FIGURE 10.53 The image of the candle is 27 cm
on the opposite side of the lens, 15 cm tall,
inverted and real.
object
convex lens
image
F
F
10.6 SAMPLE PROBLEM 2
The candle is moved so that it is now 5 cm in front of the same lens. Use ray tracing to determine the
location, size, orientation and type of image formed.
SOLUTION:
Draw the principal axis, the focal points, the object
and the three rays.
FIGURE 10.54 The image of the candle is
10 cm on the same side of the lens, 20 cm
tall, upright and virtual.
convex
lens
object
image
F
F
TOPIC 10 Ray model of light 241
10.6.4 Images formed by diverging
lenses
FIGURE 10.55 The ray diagram for a
diverging lens.
ray 1
The diverging lens shares the same features as the converging lens; however, the diverging lens can only form
Object
virtual images, because it can never bring light rays to
ray 2
focus at a point and so form a real image.
Consider an object placed outside the focus of a diverging
F Image
F
lens as shown in figure 10.55. By using ray diagrams as we
ray 3
did with converging lenses, we can observe the virtual
nature of the image formed by the diverging lens.
In the figure, our first ray approaching the lens parallel
to the principal axis is refracted away from the lens axis as it passes through so that the refracted ray
appears to come from the focus nearest the object. The second ray is directed towards the focus on the
opposite side of the lens. On reaching the lens axis, the emerging ray is directed parallel to the principal
axis. The third ray travelling from the top of the object through to the centre of the lens passes through, as
before, undiverted.
As the diagram indicates, these three refracted rays will never meet and, so, never form a real image.
Instead, they form a virtual image at the location where the three rays seem to have a common origin. The
virtual image formed here is smaller than the object, is upright and lies within the focus on the same side
of the lens as the object.
10.6 SAMPLE PROBLEM 3
The lens is switched with a diverging lens with a focal length of −10 cm. What image of the candle is
formed when it is placed 15 cm from the lens?
SOLUTION:
Draw the principal axis, focal points, object
and the three rays.
FIGURE 10.56 The image of the candle is 6 cm on
the same side of the lens, 4 cm tall, upright and
virtual.
F
object image
F
principal
axis
10.6.5 The thin lens equation
Just as the mirror equations allow more precise evaluations of image size and position than those provided
by ray tracing, so too can the thin lens equations allow the calculation of the position and size of the images
formed by lenses.
Look at figure 10.57. The two triangles shaded in green are similar triangles as all of their corresponding
angles are the same size. This would be true wherever the image is located.
242 Jacaranda Physics 11
This means that the ratios of
equivalent sides are equal, for
example:
H0 u − f
=
H1
f
Also, the triangles shaded in
blue are similar:
H0 u
=
H1 v
FIGURE 10.57
Ho
F
F
I
O
H1
u
f
f
v
The left-hand sides of these equations are equal, so we can say:
u u−f
=
v
f
Which is the same as:
u u
= −1
v f
1 1 1
= −
v f u
1 1 1
= +
f u v
This formula is known as the thin lens formula. It gives a good approximation for thin lenses. When using
it, you need to be careful with signs:
•• f is positive for converging lenses and negative for diverging lenses
•• u is positive
•• v is positive when the image is on the opposite side of the lens to the object and negative when on the
same side
We can compare the results of this formula with what we determined by ray tracing back in 10.6 Sample
problem 1.
10.6 SAMPLE PROBLEM 4
Use the thin lens formula to find the position of the image when f = 10 cm and u = 16 cm.
SOLUTION:
1 1 1
= +
u v
f
1 1 1
= −
v
f u
1
1
1
=
−
v 10 16
8
5
1
=
−
v 80 80
3
1
=
v 80
80
v =
= 27 cm behind the lens.
3
TOPIC 10 Ray model of light 243
10.6 SAMPLE PROBLEM 5
Use the thin lens formula to find the position of the image when f = 10 cm and u = 5 cm.
SOLUTION:
1
f
1
v
1
v
1
v
1
v
v
=
=
=
=
=
=
1 1
+
u v
1 1
−
f u
1
1
−
10 5
1
2
−
10 10
1
−
10
−10 cm
The image is 10 cm in front of the lens (on the same side as the object).
We have defined the magnification to be equal to the height of the image divided by the height of the
H
H
v
object, M = 1. We can see from the similar triangles in our derivation of the thin lens formula that 1 = .
u
H0
H0
H1
H1
is negative . Conversely, if v is negative, the image is upright
is positive .
However, zinverted
H0
H0
v
To account for this we add a negative to the formula so that we have M = − .
u
(
)
(
WORKING SCIENTIFICALLY 10.4
Design and build a simple telescope, documenting each stage of development and construction.
WORKING SCIENTIFICALLY 10.5
Find an old pair of glasses and, through experimentation, determine the focal length of each of the lenses and
the possible defects that the glasses were meant to correct.
10.6 Exercise 1
1 The box in the following diagram contains a lens. The scale of the grid is 1 cm per line. Draw rays from the
object to the image to determine:
(a) whether the lens is converging or diverging
(b) its focal length.
FIGURE 10.58
object
image
244 Jacaranda Physics 11
)
2 A convex lens has a focal length of 12 cm and is used as a magnifying glass by placing an object 4 cm from
the lens. Determine the magnification achieved by the magnifying glass.
3 A 4 cm high object is placed 10 cm in front of a diverging lens with a focal length of 6 cm.
(a) How far from the lens will the image appear?
(b) Will the image be real or virtual?
(c) How high will the image be?
4 When a 5 cm high object is placed in front of a concave lens of focal length 8 cm, it forms an image
2.5 times smaller than the object. What is the distance between the object and the lens?
5 Optometrists and opticians describe the focusing ability of a lens in terms of its power, P, which is equal to
the inverse of the lens’s focal length f :
1
P=
f
The unit of measurement for this type of power is the dioptre (D), where f is measured in metres.
Susan has glasses with a power of – 4.0 D.
(a) What will be the focal length of her lenses?
(b) Will the lenses be converging or diverging lenses?
(c) Is Susan more likely to be short-sighted or long-sighted?
(d) An object is placed 40 cm in front of one of Susan’s lenses. Where will the image formed by the lens
appear?
10.7 Tricks of the light
10.7.1 On a bender
Many odd visual effects that you may have noticed can be explained by the refraction of light. One that we
see all the time is the way that straight objects such as drinking straws, pencils and poles appear to be bent
when placed in water. This phenomenon can be explained using the ray model.
Consider a straight pole placed into the pool in figure 10.60. Light rays from the submerged end of pole
(P) can be drawn in all directions and, when they hit the interface between the water and the air, they bend
because of refraction. As the refractive index of air is smaller than that of water, the light is bent away from
the normal, and r > i. Some of these refracted rays originating at P find their way to the observer’s eyes.
However, as these refracted rays appear to originate from a position P′, the observer sees the image of the
end of the pole here rather than in its true position. As a result, P appears to be closer to the surface than it
really is. A similar thing happens for every point along the pole. As they all appear closer, the pole appears
bent.
FIGURE 10.59 An example of
refraction.
FIGURE 10.60 A straight pole appears bent
where it enters the water.
Normal
Refracted
ray
r
P′
Incident
rays
i
P
TOPIC 10 Ray model of light 245
PHYSICS IN FOCUS
Apparent depth
Spear throwers need to aim below a fish if they are to have a chance of spearing the fish. A similar
phenomenon occurs when a spear thrower is directly above a fish. The fish appears to be closer to the
surface than it actually is. This observation is known as apparent depth. Swimming pools provide
another example of apparent depth: they look shallower than they actually are. The refraction of light
combined with our two-eyed vision makes the pool appear shallower.
The relationship is illustrated in figure 10.61 and can be expressed as follows:
real depth
= refractive index
apparent depth
FIGURE 10.61 The phenomenon of apparent depth.
air
water
apparent
depth
real
depth
10.7.2 Total internal reflection
As seen earlier in this topic, some of the light incident
FIGURE 10.62 There are no mirrors in a fish tank
on a transparent surface will be reflected, while the
but strange reflections can be seen.
rest will be transmitted into the next medium, as
shown in figure 10.63a. This applies whether the
refracted ray is bent towards or away from the normal.
As we know from Snell’s Law, an increase in the
incident angle results in an increase in the reflected
angle. However, a special situation applies when rays
travelling from a medium with a higher refractive
index into a medium with a lower refractive index
meet the interface at certain incident angles. As the
incident angle increases in size, it will reach a critical
angle, ic, at which the angle of refraction equals its
maximum value of 90o with the normal. At this point,
the refracted ray travels parallel to the boundary between the two media (figure 10.63b).
The value of the critical angle depends upon the refractive indices of the two media. At the critical angle ic:
n1sin (ic) = n2sin (90o)
so,
n1sin (ic) = n2
and thus,
n
sin (ic) = 2
n1
246 Jacaranda Physics 11
FIGURE 10.63 Total internal reflection.
(a) i < ic
(b) i = ic
(c) i > ic
refracted
ray
n1
i
incident
ray
no
refraction
refracted
ray
r
n2
i c ic
i
reflected
ray
(total internal reflection)
incident
ray
i
reflected
ray
incident
ray
i
reflected
ray
The value n2/n1 is referred to as the relative refractive index for media 1 and 2. It should be noted that a
critical angle can only exist provided that the relative refractive index is less than 1 — that is, n1 > n2.
At incident angles greater than the critical angle, all the light is reflected back into the original medium and
no refracted ray is formed. This circumstance is referred to as total internal reflection (figure 10.63c).
Total internal reflection is a relatively common atmospheric phenomenon (as in mirages) and it has
technological uses (for example, in optical fibres).
10.7 SAMPLE PROBLEM 1
What is the critical angle for light rays passing from water into air given that the refractive index of
water is 1.3?
SOLUTION:
nair = 1.0; θ air = 90°; nwater = 1.3; θ water = ?
1.3 × sin θ water = 1.0 × sin 90° (substitute data into Snell's Law)
sin 90°
(rearrange formula to get the unknown by itself)
√sin θ water =
1.3
= 0.7692 (determine sine values and calculate expression)
θ water = 50.28° (use inverse sine to find angle)
θ water = 50° (round off to two significant figures)
10.7.3 Dispersion of light
FIGURE 10.64 Dispersion by a glass prism.
Isaac Newton was the first person to discuss the
Glass Prism
breaking up of white light into the coloured spectrum
in a process called dispersion. He observed that, as
white light passes through a triangular glass prism, the
White
individual spectral colours emerge. This occurs because
Red
light
the different colours of the visible light spectrum travel
through the glass at different velocities. This means
that light of each colour has a different refractive index
Violet
and so the different colours are refracted through at
different angles. Violet light, which travels the slowest,
is refracted the most, while red light, which travels the
fastest, is refracted the least.
Rainbows are formed when sunlight is incident on
water particles suspended in the air, which is why they are most frequently seen after rain showers. When
white light from the sun enters the water droplet, it is refracted and dispersion occurs, separating the individual colours. These rays continue to travel until reaching the far surface of the water droplet, where some
TOPIC 10 Ray model of light 247
emerge but the rest are totally internally reflected back into
FIGURE 10.65 Dispersion in a water droplet.
the droplet. When they encounter the front boundary
Refraction
between the water of the droplet and the air, the rays are
White ligh
t
Dispersion
1 from the su
again refracted, further increasing the angle of dispersion.
n 2
The only dispersed rays reaching our eyes from each of
3 Total internal
Rain
those billions of droplets are those that have an angle
reflection
drop
of spectrum
between 40o (violet light) and 42o (red light) relative to the
Red
incident sunlight.
4
The arc of colour seen by the observer is the section of a
Refraction
et
Viol
circle subtended by these angles at a point called the antisolar point, which lies on the line between the observer’s
Rainbow
eye and the sun.
The higher above the horizon the observer is, the higher above the ground the antisolar point is positioned and so the greater the proportion of the circle that is seen. At a very high altitude, an entire circular
rainbow could be observed.
FIGURE 10.66 Angle of dispersion.
Sunlight
FIGURE 10.67 How the arc of a rainbow forms.
42˚
sunlight
42˚ 40˚
Rain drop
observer
antisolar
point
10.7 Exercise 1
1 Which of the following colours of light travels the fastest through glass:
(a) blue
(b) green
(c) yellow
(d) violet?
2 A glass fibre has a refractive index of x and its cladding has a refractive index of y. What is the critical angle
in the fibre?
3 What is the critical angle for light passing from diamond into water?
4 The critical angle for light passing from a mystery liquid into air is 43.2o. What is the absolute refractive
index of the mystery liquid?
5 A light positioned in the bottom of a 1.5 m pool produces a circle of light on the water’s surface. What is the
radius of the light circle?
6 Mark stands at the edge of a fish pond and sees a large fish in the water. From where he is standing, the
fish is 2 m horizontally from the pond’s edge and appears to be 50 cm below the surface. How far below the
surface of the pond is the fish actually located? Assume that Mark’s eyes are 1.5 m above pond level.
7 Phuong placed a coin in the bottom of an opaque mug. From where she is sitting, she can’t quite see the
coin. However, when she pours some water into the mug, she finds that she can now see the coin. How is
this possible?
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248 Jacaranda Physics 11
10.8 Review
10.8.1 Summary
•• The ray model depicts light as straight lines in a uniform medium.
•• All electromagnetic waves travel at the same speed in a vacuum and are slowed down when they enter
any other media.
•• The speed of light in a vacuum is 299 792 458 m s–1, usually approximated to 3 × 108 m s–1.
•• A luminous body is one that can directly produce light. A body that produces light when heated is said
to be incandescent. A non-luminous or illuminated body is one that does not itself produce light, but
reflects it from another source of light.
•• The incident ray, reflected ray and the normal to the surface all lie in the same plane.
•• The absolute refractive index of a transparent medium is the ratio of the speed of light in a vacuum to the
speed of light in the medium. The refractive index is always larger than 1.
•• A transparent material is one through which an object may be clearly seen. A translucent material allows
light through it, but does not allow an object to be seen coherently through it. An opaque material is one
through which light cannot pass at all.
•• A material may reflect, transmit or absorb light, or a combination of these, depending upon the nature of
the material.
•• The Law of Reflection: the angle of incidence is equal to the angle of reflection.
•• A concave (converging) mirror reflects parallel light rays so that they converge on the focal plane of the
mirror. A convex (diverging mirror) reflects parallel light rays so that they spread out.
•• Ray tracing and the mirror equations can be used to determine the location, size and nature of images
produced by curved mirrors.
•• Light is refracted when it passes between different transparent materials. The degree of refraction is
described by Snell’s Law: n1 sin i = n2 sin r.
•• A lens is a device made from a transparent medium that allows the refraction of light to be controlled.
•• A converging lens is thicker in the middle than at the edges. Parallel rays passing through a converging
lens coincide at the focus of the lens. A diverging lens is thicker at its edges than in its middle. Parallel
rays passing through a diverging lens spread out so that they appear to originate at a point on the focal
plane nearest the object.
•• The focal length of a lens depends upon the curvature of the faces and the refractive index of the medium
from which it is made.
•• The object distance (u) for lenses is assumed to be positive. The image distance (v) is negative for virtual images and positive for real images.
•• A converging lens has a positive focal length, f , while a diverging lens has a negative focal length.
•• A real image is one created from converging light rays. It will manifest on a screen placed at the formation position.
•• A virtual image is unable to materialise on a screen, and can only be seen when viewed in a mirror or
through a lens. Light rays do not converge at a virtual image.
•• The position of an image formed by thin lenses can be determined by accurate ray tracing and by using
1 1 1
the thin lens equation: = + .
f u v
10.8.2 Questions
1. Which of the following are luminous objects:
(a) the Sun
(d) a campfire
(b) a projector screen
(e) a red-hot piece of iron
(c) a star
(f) a television screen?
TOPIC 10 Ray model of light 249
2. True or false? One type of transparent plastic has a refractive index equal to that of water (n = 1.33) .
If you placed a lump of this plastic into water, you would not see it.
3. Calculate the angles a, b and c in figure 10.68.
FIGURE 10.68
4. What is the luminous intensity of a 50 W light bulb at a distance of 2 m
if we assume that all of the bulb’s energy is converted into light?
5. A light source with a luminous intensity of 36 W m–2 that is positioned
1.2 m from a light meter produces the same reading as a second light
mirror
source that is positioned 2.4 m away from the meter. What is the lumic
50°
nous intensity of the second source?
a b
6. What is the angle of refraction in water (n = 1.33) for an angle of
incidence of 40°? If the angle of incidence is increased by 10°, by how
much does the angle of refraction increase?
7. A ray of light enters a plastic block at an angle of incidence of 55° with an angle of refraction of 33°.
What is the refractive index of the plastic?
8. A ray of light passes through a rectangular glass block with a refractive index of 1.55. The angle of
incidence as the ray enters the block is 65°. Calculate the angle of refraction at the first face of the
block, then calculate the angle of refraction as the ray emerges on the other side of the block. Com
ment on your answers.
9. Immiscible liquids are liquids that do not mix. Immiscible liquids will
FIGURE 10.69
settle on top of each other, in the order of their density, with the densest
liquid at the bottom. Some immiscible liquids are also transparent.
light
ray
(a) Calculate the angles of refraction as a ray passes down through
25˚
air
immiscible layers as shown in figure 10.69.
n = 1.00
n = 1.357
(b) If a plane mirror was placed at the bottom of the beaker,
acetone
n = 1.4746
calculate the angles of refraction as the ray reflects back to the
glycerol
n = 1.4601
carbon
surface. Comment on your answers.
tetrachloride
10. Light rays are shown passing through boxes in figure 10.70. Identify
n = 1.53
glass beaker
the contents of each box from the options (a)–(g) given below.
Option (b) is a mirror. All others are solid glass.
Note: There are more options than boxes.
FIGURE 10.70
(ii)
(i)
(a)
(b)
(iii)
(c)
(iv)
(d)
(v)
(e)
(vi)
(f)
(g)
11. An object is placed 40 cm in front of a convex mirror that has a focal length of 30 cm.
(a) Where is the image formed?
(b) Is the image:
i real or virtual,
ii reduced or enlarged, or
iii inverted or upright?
12. A concave mirror has a 40 cm focal length. How far from the mirror must an object be positioned in
order for:
(a) an image to appear 50 cm from the mirror
(b) a real image to be formed that is twice the height of the object
(c) a virtual image to be formed that is three times the height of the object?
250 Jacaranda Physics 11
13. Use ray tracing to determine the full description of the following objects:
(a) a 4.0 cm high object, 20 cm in front of a convex lens with a focal length of 15 cm
(b) a 3.0 mm high object, 10 cm in front of a convex lens with a focal length of 12 cm
(c) a 5.0 cm high object, 200 cm in front of a convex lens with a focal length of 10 cm.
14. What does ‘accommodation mechanism’ mean? Give an example.
15. (a)You are carrying out a convex lens investigation at a bench near the classroom window and you
obtain a sharp image of the window on your screen. A teacher walks past outside the window.
What do you see on the screen?
(b) The trees outside the classroom are unclear on the screen. What can you do to bring the trees into
focus?
16. Use ray tracing to determine the magnification of an object placed under the following two-lens
microscope. The object is placed 5.2 mm from an objective lens of focal length 5.0 mm. The eyepiece
lens has a focal length of 40 mm. The poles of the lenses are 150 mm apart.
17. A convex lens with a focal length of 5.0 cm is used as a magnifying glass. Determine the size and
location of the image of text on this page if the centre of the lens was placed:
(a) 4.0 cm above the page
(b) 3.0 cm above the page.
18. A 35 mm slide is placed in a slide projector. A sharp image is produced on a screen 4.0 m away. The
focal length of the lens system is 5.0 cm.
(a) How far is the slide from the centre of the lens?
(b) What is the size of the image?
(c) Looking from the back of the slide projector, the slide contains a letter ‘L’. What shape will appear
on the screen?
(d) The slide projector is moved closer to the screen. The image becomes unclear. Should the lens
system be moved closer to or further away from the slide?
19. An object is placed at an equal distance a from two plane mirrors
FIGURE 10.71
that are placed at right angles as shown in figure 10.71. How
Mirror 1
many images of the object are formed in the mirrors?
20. (a)What is the angle of refraction in water (n = 1.33) of a light
ray that has an incident angle of 30°?
a
(b) By how much will the angle of refraction increase if the
Mirror 2
incident angle is increased by 10°?
21. Calculate the distance the photographic film needs to be from the
a
centre of a camera lens of focal length 5.0 cm in order to take a
sharp, focused photograph of a family group located 15.0 m away.
FIGURE 10.72
Extra lenses here
Film winding
sprocket
lens
Movement
Axis
Shutter
Shutter
moves
Object
Converging lens
Pressure plate
Gaps in
shutter
Iris diaphragm
Aperture ring
Film take-up
spool
Film
Film cassette
Focusing ring
TOPIC 10 Ray model of light 251
di (cm)
22. Calculate the sideways deflection as a ray of light goes through a parallel-sided
FIGURE 10.73
plastic block (n = 1.4) with sides 5.0 cm apart, as in figure 10.73.
23. During the course of an experiment, a student moves an object into
30°
different positions (do) in front of a converging lens and measures the
resulting positions (di) of the image formed. His results are then plotted
on a graph as shown in figure 10.74.
n = 1.4
5 cm
(a) What is the focal length of the lens?
(b) Describe the image formed for point X. Is it real or virtual? Is it
enlarged or reduced?
(c) If the object is placed at the 50 cm position, where will the image
form?
24. A lens held 20 cm from an object produces a real, inverted
FIGURE 10.74
image of it 30 cm on the other side of the lens. What is the
60
50
focal length of the lens? Is it converging or diverging?
40
25. A convex air pocket is formed inside a block of Perspex as
30
shown in figure 10.75. What effect will this air pocket have
20
10
on parallel light rays entering the block?
0
d (cm)
40
50
60
10
20
30
26. A converging lens with a focal length of 40 cm is placed 1 m in
–10
–20
front of a diverging lens that has a focal length of 30 cm. A
–30
5 cm birthday candle is lit and placed a distance of 80 cm in
x
–40
–50
front of the converging lens, as shown in figure 10.76. Under
the influence of both lenses, where will the final image be
formed, and how high will it be?
FIGURE 10.75
27. Describe the light path from a light source to your eye in seeing an
object.
Air gap
Perspex
28. Use the ray model and the sources of light to rephrase the
statements (a) ‘I looked at a flower through the window’ and
(b) ‘I watched the TV’.
29. Explain how early astronomers knew the Moon must have a rough
surface.
30. Copy figure 10.77 and draw the incident and reflected rays from
the two ends of the object to the eye. Locate the image.
FIGURE 10.76
31. The two arrowed lines in figure 10.78 represent reflected rays. The line
Converging lens Diverging lens
AB represents the plane mirror. Locate the image and the light source in
each of the two figures.
o
5 cm
FIGURE 10.78
A
A
80 cm
1m
FIGURE 10.77
B
B
object
Plane
mirror
32. A student argues that you cannot photograph a virtual image because
light rays do not pass through the space where the image is formed. How
would you argue against this statement?
252 Jacaranda Physics 11
33. Sketch the path of the rays entering
FIGURE 10.79
each of the pair of joined mirrors in
figure 10.79.
34. When slides are placed in a slide projector, they are put into the cartridge
upside-down. Why is this done?
35. Explain how you can use two mirrors
to see your view from behind.
36. As part of an experiment on Snell’s
Law, a student measures the angle of
refraction (θ 2) obtained when an
incident light ray enters a clear plastic block for a number of different θ 1 (degrees)
incident angles (θ 1). Her results are shown in the table at right.
0
(a) Draw a graph of this data.
10
(b) Use your graph to determine the refractive index of the plastic.
20
37. When you look into a plane mirror, your left and right sides appear
30
reversed. This is called lateral inversion. Draw a diagram showing
40
how you could position a series of plane mirrors so that you see your
50
image upside-down.
θ 2 (degrees)
0
6
12
18
24
29
PRACTICAL INVESTIGATIONS
Investigation 10.1: Snell’s Law
Aim
To observe the refraction of light and to use Snell’s Law to determine the refractive index of a medium
Materials
Power supply, ray box with single slit card, rectangular Perspex or glass block, ruler, protractor, pencil, blank A4
paper, drawing board, drawing pins
Method
1. Use drawing pins to attach the A4 paper to the drawing board, which should be lying flat on the bench.
2. Place the block in the middle of the page. Use a pencil to draw around the block so that it can always be
returned to the same position. Mark a point on the boundary and label it as O.
3. Reduce the amount of light in the room (by drawing curtains etc.). Turn on the ray box and direct a single ray
of light so that it enters the block at point O at an angle and emerges on the other side of the block.
4. Without moving the ray or the block, mark 3 points along each of the incident ray and the emerging ray.
Place a mark at the block boundary at the point where the light ray emerges from the block and label this R.
5. Turn off the ray box and remove the block from the paper. Using your pencil marks as guides, use a ruler to
draw the path of the incident ray into the block, joining points O and R, and to draw the path of the emerging
ray. Draw normals to the surface at points O and R.
6. Use your protractor to measure the angle of incidence θ 1 and the angle of refraction θ 2 as shown in
figure 10.80. Enter these values into table 10.4A.
7. Repeat steps 1–6 for three other incident angles.
Results
TABLE 10.4A
θ 1 (degrees)
θ 2 (degrees)
TOPIC 10 Ray model of light 253
Analysing the results
1. For each of the angles in table 10.4A, complete table 10.4B
at right.
2. Snell’s Law states that n1 sin θ 1 = n2 sin θ 2. Given that
sin θ 2
n1 = 1.00 (air), what variable does the ratio
sin θ 1
represent?
3. What is the approximate refractive index of the block?
4. In each case, is the ray refracted towards the normal or
away from the normal as it passes into the block?
5. What do you notice about the angles of the incident ray
entering the block at O and the ray emerging from the
block at R?
TABLE 10.4B
sin θ 1
sin θ 2
sin θ 2
sin θ 1
FIGURE
10.80
R
Conclusion
State the relationship between the refractive index of the block, the
angle of incidence and the angle of refraction in this investigation.
Investigation 10.2: Concave mirrors — an observation
exercise
𝜃2
This investigation involves observing yourself in a concave mirror.
O
𝜃1
You will need the following equipment:
• concave mirror
• tape measure or metre ruler.
Look at yourself in a concave mirror.
1. How does your appearance change as you move towards and away from the mirror?
2. Describe your image (for example, size and orientation) as the distance changes. Note the distance.
3. Was there a distance at which the image changed markedly? If so, where did you notice that this occurred?
4. Do you notice any distortion of the image? If so, how was the image distorted and where did this occur?
Investigation 10.3: Converging Lenses
Aim
To investigate the formation of images by converging lenses
Materials
Biconvex glass lens, lens holder, metre ruler, small birthday candle mounted in a holder, white cardboard
screen, masking tape
Method
1. Place the metre ruler flat on the benchtop.
2. Put the biconvex lens in the holder and place it next to the ruler at the 50 cm mark.
3. Light the candle and place it next to the 0 cm mark of the ruler. This location corresponds to do = 50 cm.
Place the screen against the 100 cm mark and move it closer to the lens or further away from the lens until a
clear image of the candle flame appears on the screen. Measure the distance di between the lens and the
screen, and enter this value into table 10.5A. The image formed on the screen is said to be a real image.
FIGURE 10.81
Candle
Lens
do
Ruler
Screen
di
4. Continue to move the candle closer to the lens in 5 cm increments and measure the corresponding values of di.
254 Jacaranda Physics 11
5. Eventually, you will reach values of do at which no clear image can be formed on the screen. This will occur
when the candle is located at the focus of the lens (where no image of the candle can be formed) or closer
(where only a virtual image can be formed). In the case of a virtual image, looking directly through the lens
itself at the candle will reveal its image. The location of the virtual image can be estimated by using the
adjacent ruler. Enter the di values of these images as negative values in table 10.5A.
Results
Analysing the results
1. When the candle is at the focus of the
lens, F, no image (either real or virtual)
can be formed. At what value of do
did this occur?
2. Using your values for do and di and the
1
1
1
+ , determine the
equation =
f do di
value of f for the lens.
TABLE 10.5A
Image
(do) cm (di) cm
3. How do your values for question 1
and question 2 compare?
4. Converging lenses are used as
magnifying glasses. Is this magnified
image a real or virtual image? Justify
your answer.
5. The power P (in dioptres) of a lens is
equal to the inverse of its focal length
(in metres). What is the power of the
lens you have used here?
Real or
virtual?
Erect or
inverted?
Enlarged or
reduced?
50
45
40
35
30
25
20
15
10
5
Conclusion
1. Complete the following table summarising
the locations and types of images formed by
a converging lens.
TABLE 10.5B
Position of object
Position of image
Description of
image
2f > do > f
do = f
f > do
Investigation 10.4: Using apparent depth to determine the refractive index
This investigation involves using apparent depth to determine the refractive index.
You will need the following equipment:
• rectangular glass or perspex block
• two pieces of grid graph paper.
Place the smallest face of a rectangular glass block on a sheet of grid graph paper as shown in figure 10.82.
TOPIC 10 Ray model of light 255
FIGURE 10.82
Look down a vertical face of the block so that you can see the grid of the graph paper through the glass and
through the air. The grid seen through the glass will appear larger (closer).
Slowly bring another piece of identical graph paper up that face until the graph pattern seen through the block
matches the pattern held beside the block. Take care, this is a difficult task.
Mark the point where the patterns are seen to match and measure its distance from the top of the block.
Repeat this exercise several times and calculate the average of your measurements.
Measure the full length of the glass block and calculate the refractive index of the block, using the equation:
real depth
apparent depth
= refractive index.
Repeat the exercise to calculate the refractive index of water, using a fish tank or a large beaker instead of the
glass block.
Investigation 10.5: Floating coins
AIM
To investigate the effect of refraction on the image of a submerged object
You will need:
2 beakers
evaporating dish
FIGURE 10.83 The image
coin
of the coin is not in the
• Place a coin at the bottom of an empty beaker and look at it from above
same place as the actual
while your partner slowly adds water from another beaker.
coin.
• Place the coin in the centre of an evaporating dish and move back just far
enough so you can no longer see the coin. Remain in this position while your
partner slowly adds water to the dish.
• Make a copy of the diagrams shown in figure 10.83. Use dotted lines to
extend back the rays shown entering the observer’s eye to see where they Beaker
seem to be coming from. This enables you to locate the centre of the image
Water
of the coin.
Discussion
Coin
1. How does the position of the coin appear to change while the water is being
Evaporating
added?
dish
2. Which other feature of the coin appears to change?
3. What appears to happen to the coin as water is added to the evaporating
dish?
Water
4. Is the image of the coin above or below the actual coin?
Coin
256 Jacaranda Physics 11
Investigation 10.6: Total Internal reflection
Aim
To observe total internal reflection of light in a Perspex prism
You will need:
ray box kit
12 V DC power supply
Perspex triangular prism
• Connect the ray box to the power supply. Place the ray box over a
FIGURE 10.84 Observe the
page of your notebook. Use one of the black plastic slides in the ray
beam of light as it passes
box kit to produce a single thin beam of light that is clearly visible on
through the prism.
the white paper.
• Place a Perspex triangular prism on your notebook and direct the thin Narrow beam
beam of light towards it as shown in figure 10.84. Observe the beam of light from
Perspex
ray box
prism
as it passes through the prism.
• Turn the prism slightly anticlockwise, closely observing the thin light
beam as it travels from the Perspex prism back into the air. Continue
to turn the prism until the beam no longer emerges from the prism.
Discussion
1. Describe what happens to the thin light beam as it passes from air into
the Perspex prism and back into the air.
2. Outline what happens to the beam of light when it no longer emerges
from the prism.
3. Draw a series of two or three diagrams showing how the path taken by
the beam of light changed as you turned the prism.
TOPIC 10 Ray model of light 257